Introduction to Trigonometry: Table Values, Ratios & Proofs Class 10

In Hellenistic astronomy, ancient surveyors and navigators measured the distances between distant stars and earthly mountains by calculating the ratios of right-angled triangles. Today, trigonometry forms the foundation of aerospace engineering, computer graphics, satellite GPS tracking, and architectural mechanics.

In CBSE Class 10 Mathematics, Chapter 8 (Introduction to Trigonometry) introduces the six trigonometric ratios: $\sin, \cos, \tan, \cot, \sec,$ and $\csc$. Board examinations frequently test two major areas: calculating and applying the trigonometric values of standard angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$), and solving simultaneous algebraic systems like $\tan(A + B) = \sqrt{3}$ to find unknown angles $A$ and $B$.

In this master guide, we derive the exact table values geometrically and work through high-scoring board exam problems.

What You Will Learn

• Definitions of the Six Trigonometric Ratios in a right-angled triangle
• Geometric derivation of trigonometric values for $45^\circ$ and $30^\circ / 60^\circ$
• The Finger-Counting Memory Trick for the $0^\circ - 90^\circ$ table
• Evaluating complex multi-term trigonometric arithmetic expressions
• Solving simultaneous trigonometric equations for unknown acute angles $A$ and $B$
• Core quotient and reciprocal relationships

1. The Six Trigonometric Ratios

In a right-angled triangle $\Delta ABC$ right-angled at $B$, taking acute angle $\theta = \angle A$:

                                  C
                                  |                                   |                     Opposite Side   |     \  Hypotenuse
                  (Perpendicular) |                                         |   θ                                       B----------A
                                  Adjacent Side (Base)

$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}, \quad \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}, \quad \tan \theta = \frac{\text{Perpendicular}}{\text{Base}}$ $\csc \theta = \frac{1}{\sin \theta}, \quad \sec \theta = \frac{1}{\cos \theta}, \quad \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$

2. Geometric Derivation of Trigonometric Table Values

Derivation for $45^\circ$ (Isosceles Right Triangle):

Consider an isosceles right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$ and $AB = BC = a$.

• Since $AB = BC$, the acute angles are equal: $\angle A = \angle C = \mathbf{45^\circ}$.
• By the Pythagoras Theorem: $\text{Hypotenuse } AC = \sqrt{a^2 + a^2} = \sqrt{2a^2} = \mathbf{a\sqrt{2}}$
• Therefore: $\mathbf{\sin 45^\circ = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}}$ $\mathbf{\cos 45^\circ = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}}$ $\mathbf{\tan 45^\circ = \frac{BC}{AB} = \frac{a}{a} = 1}$

Derivation for $30^\circ$ and $60^\circ$ (Equilateral Triangle Altitude):

Consider an equilateral triangle $\Delta ABC$ with side length $2a$. Draw altitude $AD \perp BC$.

• In $\Delta ABC$, $\angle A = \angle B = \angle C = 60^\circ$.
• The altitude bisects $\angle A$ into two $30^\circ$ angles and bisects $BC$ such that $BD = a$.
• In right triangle $\Delta ABD$: $AD = \sqrt{(2a)^2 - a^2} = \sqrt{4a^2 - a^2} = \mathbf{a\sqrt{3}}$
• For $\angle B = \mathbf{60^\circ}$: $\sin 60^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \mathbf{\frac{\sqrt{3}}{2}}, \quad \cos 60^\circ = \frac{BD}{AB} = \frac{a}{2a} = \mathbf{\frac{1}{2}}, \quad \tan 60^\circ = \mathbf{\sqrt{3}}$
• For $\angle BAD = \mathbf{30^\circ}$: $\sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \mathbf{\frac{1}{2}}, \quad \cos 30^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \mathbf{\frac{\sqrt{3}}{2}}, \quad \tan 30^\circ = \mathbf{\frac{1}{\sqrt{3}}}$

3. The Master Trigonometric Values Table

Important: <u>Notice how $\cos heta$ values are simply the $\sin heta$ row written in reverse order! Furthermore, $an heta = \sin heta / \cos heta$, so memorizing the single sine row gives you the entire trigonometric table!</u>

4. High-Yield Solved Board Examination Problems

Solved Example 1: Evaluating Multi-Term Arithmetic Expression

Problem: Evaluate: $\frac{5\cos^2 60^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

Solution:

1. Recall the Fundamental Identity: In the denominator: $\sin^2 30^\circ + \cos^2 30^\circ = \mathbf{1}$!
2. Substitute Table Values into the Numerator:
   • $\cos 60^\circ = \frac{1}{2} \implies \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$
   • $\sec 30^\circ = \frac{2}{\sqrt{3}} \implies \sec^2 30^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}$
   • $\tan 45^\circ = 1 \implies \tan^2 45^\circ = 1^2 = 1$
3. Compute the Expression: $\text{Value} = \frac{5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) - 1}{1} = \frac{5}{4} + \frac{16}{3} - 1$
4. Take Common Denominator (LCM of 4 and 3 is 12): $\text{Value} = \frac{5(3) + 16(4) - 1(12)}{12} = \frac{15 + 64 - 12}{12} = \frac{79 - 12}{12} = \mathbf{\frac{67}{12}}$
5. Therefore, <u>the value of the expression is rac{67}{12}</u>.

Solved Example 2: Finding Unknown Acute Angles A and B (CBSE Classic)

Problem: If $\tan(A + B) = \sqrt{3}$ and $\tan(A - B) = \frac{1}{\sqrt{3}}$, where $0^\circ < A + B \le 90^\circ$ and $A > B$, find the values of $A$ and $B$.

Solution:

1. Given: $\tan(A + B) = \sqrt{3}$. Since $\tan 60^\circ = \sqrt{3}$: $\mathbf{A + B = 60^\circ} \quad \text{--- (Equation 1)}$
2. Given: $\tan(A - B) = \frac{1}{\sqrt{3}}$. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$: $\mathbf{A - B = 30^\circ} \quad \text{--- (Equation 2)}$
3. Add Equation 1 and Equation 2: $(A + B) + (A - B) = 60^\circ + 30^\circ$ $2A = 90^\circ \implies \mathbf{A = 45^\circ}$
4. Substitute $A = 45^\circ$ into Equation 1: $45^\circ + B = 60^\circ \implies B = 60^\circ - 45^\circ = \mathbf{15^\circ}$
5. Therefore, <u>the values are $A = 45^\circ$ and $B = 15^\circ$</u>.

5. Summary and Examination Tips

Exam Tip: In questions where the denominator has $\sin^2 \theta + \cos^2 \theta$, immediately replace it with $1$! Do not waste time computing $(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{1}{4} + \frac{3}{4} = 1$.

Common Mistake: Writing $\tan(A+B) = \tan A + \tan B$. Trigonometric functions do NOT obey the distributive law over addition! $\tan(A+B) \ne \tan A + \tan B$.